Stripe·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Stripe coding round for a software engineer role, one problem the whole session. The problem looked like a clean DP question on the surface but the skip mechanic made it messier than expected.

Questions Asked (1)

You have N factories in a sequence, each with multiple options described by a cost and a distance. You must pick exactly one option per factory you decide to build, and you must skip exactly one factory entirely. The goal is to minimize the total selected costs plus the sum of absolute differences between consecutive built factories' distances (the skipped factory is removed from the sequence, so its two neighbors become adjacent). How do you solve this, and what's the complexity?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was to just try every possible skip index, which is N candidates, and for each one run a DP over the remaining N-1 factories.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem constraints and edge cases (e.g., N≥3, costs and distances ranges). Then, propose a dynamic programming solution that tracks the last built factory and whether a skip has been used, using prefix/suffix minima to handle the skip efficiently. Finally, analyze the time and space complexity, and discuss potential optimizations or trade-offs.

Pro tip: Mention that the skip can be handled by precomputing prefix and suffix DP arrays, which is a common pattern in sequence DP problems and shows you can optimize beyond naive O(N^2).

1. Clarify the problem

Restate the problem to ensure understanding: exactly one factory is skipped, one option per built factory, minimize sum of costs plus absolute distance differences between consecutive built factories. Ask about constraints (N, number of options, value ranges) and edge cases (e.g., N=3, all costs positive).

2. Define DP state

Define DP[i][j][k] where i is the current factory index, j is the chosen option index for factory i (if built), and k is a boolean indicating whether a skip has been used. Alternatively, define DP[i][k] as the minimum cost up to factory i with skip status k, but need to track last built option for distance calculation.

3. Formulate transitions

For each factory, consider building it (choose an option) or skipping it (if skip not used). When building, add option cost plus distance difference from the last built factory's option. When skipping, carry forward the last built state without adding distance.

4. Optimize with prefix/suffix

To avoid O(N^2) due to distance differences, precompute prefix DP up to each factory and suffix DP from each factory, then combine at the skipped factory. This reduces complexity to O(N * M^2) where M is max options per factory, or O(N * M) with further optimization.

5. Analyze complexity and trade-offs

State time and space complexity: O(N * M^2) time and O(N * M) space for the DP, or O(N * M) time with prefix/suffix minima. Discuss trade-offs: memory vs. time, and whether the solution scales for large N and M.

Key Points to Mention

Dynamic programming with state tracking the last built factory and skip usage.
Handling the skip by splitting the sequence into prefix and suffix, then combining.
Using prefix/suffix minima to optimize distance difference calculations.
Time complexity: O(N * M^2) naive, O(N * M) optimized; space complexity O(N * M).
Edge cases: N=3 (must skip one, build two), all options have same distance, negative costs?
Trade-offs: memory vs. time, and potential for further optimization if M is large.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.